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Bunuel
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subhajit1
Hi friends,

the line y=2x surely passes through the origin as the coordinates (0,0) satisfies the equation.

Now coming to the point, the solution says both the statements are required, but if you take the slope of second line as -1/2 then the equation for is

-1/2=(y-b)/(x-a)
or, 2y= -x +(a+b)
or, y=(-1/2)x+(a+b)/2

Now, applying statement 1 (a=-b)to this equation;

y=(-1/2)x+{a+(-a)}/2
or, y=(-1/2)x+ 0/2
or, y=(-1/2)x

So, statement 1 seems sufficient (contradiction with the actual answer)

Check the diagram below:



As you can see if a = 1 and b = -1 we have different line perpendicular to y = 2x than if a = 2 and b = -2.

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Hi, Can anyone tell me what will be the final equation of the line once we know what a and b are?
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Hi, Can anyone tell me what will be the final equation of the line once we know what a and b are?

From a=-b and a-b=1, we can get that a = 1/2 and b = -1/2. So, we know that k passes through the point (1/2, -1/2) and is perpendicular to line y = 2x (slope = 2)

The two lines are perpendicular if and only if the product of their slopes is -1, so m*2 = -1 and m = -1/2 (the slope of line k).

Finally, the equation of a straight line that passes through a point \(P_1(x_1, y_1)\) with a slope m is: \(y-y_1=m(x-x_1)\). Substitute: \(y-(-\frac{1}{2})=-\frac{1}{2}(x-\frac{1}{2})\) --> \(y = -\frac{x}{2} - \frac{1}{4}\).

Check below:




For Coordinate Geometry check:

24. Coordinate Geometry



For more check:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread

Hope it helps.

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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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